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\maketitle

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%\thispagestyle{empty} % 第一页不显示页码

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\section{Fundamental Theorems } 

\begin{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %%1
(***) 
Let $p(x,y)$ and $q(x,y)$ be continuous functions in a region $\Omega$. 
The line integral $$\int_\gamma p dx + q dy,$$ defined in $\Omega$, depends only on the end points of $\gamma$ if and only if there exists a function $U(x,y)$ in $\Omega$ with the partial derivatives $\partial U/\partial x = p$ and $\partial U/\partial y= q.$

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 

\begin{enumerate}[label={\arabic*.}]

\item  Proof of sufficiency. 
 
\begin{enumerate}[label={\arabic*.}]
\item  Suppose there is a function $U(x,y)$ in $\Omega$ with $\partial U/\partial x = p$ and $\partial U/\partial y= q$. 
\item  Let the curve $\gamma$ be defined by $z=z(t), a\le t\le b$. 
\item  We can compute the line integral explicitly, 
\begin{equation*}
\begin{aligned}
\int_\gamma p dx + q dy &= \int_\gamma \left( \frac{\partial U}{\partial x} dx + \frac{\partial U}{\partial y} dy  \right) 
 = \int_a^b \left( \frac{\partial U}{\partial x} x'(t) + \frac{\partial U}{\partial y} y'(t)  \right)dt 
 = \int_a^b \frac{dU}{dt}dt \\ 
&= U(x(b),y(b)) - U(x(a),y(a)). 
\end{aligned}
\end{equation*}

\item  The integral depends only on the end points of $\gamma$. 

\end{enumerate}


\item  Proof of necessity. 
 
\begin{enumerate}[label={\arabic*.}]
\item  We choose a fixed point $(x_0,y_0)$ in $\Omega$. 

\item  Since the line integral depends only on the end points of $\gamma$, we can define a function 
$$
U(x,y) = \int_\gamma pdx+qdy, 
$$
where $\gamma$ is any polygon contained in $\Omega$, connecting $(x_0,y_0)$ and $(x,y)$. 

\item  To compute $\frac{\partial U}{\partial x}$, we choose the last segment of $\gamma$ horizontal. 
Then 
$$
\frac{\partial U}{\partial x} = \lim\limits_{h\to 0} \frac{U(x+h,y)-U(x,y)}{h} 
= \lim\limits_{h\to 0} \frac{1}{h} \int_{(x-h,y)}^{x,y} p(x,y)dx = p(x,y). 
$$

\item  Similarly we can choose another path and verify that $\frac{\partial U}{\partial y}=q(x,y)$. 

\end{enumerate}

\end{enumerate}

} % End of Solution. 

\fi

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%\newpage
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\item  %%2
Compute the complex line integral 
$$\int_\gamma xdz$$ 
where $\gamma$ is the directed line segment from $0$ to $1+i$.

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  Let the arc $\gamma$ be defined by $z(t)=(1+i)t$, where $0\le t\le 1$. 
\item  The definition of complex line integral is 
$$
\int_\gamma f(z)dz = \int_a^b f(z(t))z'(t)dt,
$$
 where $z(t),a\le t\le b$ is the equation of the piecewise differentiable arc $\gamma$. 
 
 \item  In this problem, we compute 
$$
\int_\gamma xdz = \int_0^1 t(1+i)dt = (1+i) \int_0^1 tdt = \frac{1+i}{2}. 
$$

\end{enumerate}

} % End of Solution. 
\fi

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%\newpage
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\item  %%3
(***) 
Compute the complex line integral 
$$\int_{|z|=r} xdz,$$ 
for the positive sense of the circle, in two ways: first, by use of a parameter, and second, by observing that 
$$ x=\frac{1}{2}(z+\bar{z})=\frac{1}{2}\left(z+\frac{r^2}{z}\right)$$ on the circle. 


\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
 
 {First Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  Use a parameter $t$, and let $z=re^{it}, 0\le t\le 2\pi$. 

\item  The Euler's formula is $e^{it} = \cos t + i\sin t$. 

\item  Compute the integral with parameter $t$, 
\begin{equation*}
\begin{aligned}
\int_{|z|=r} xdz &= \int_0^{2\pi} (r\cos t) (re^{it}idt) = ir^2 \int_0^{2\pi} (\cos^2t + i\cos t \sin t) dt \\ 
&= ir^2 \int_0^{2\pi} \left( \frac{1+\cos 2t}{2} + i\frac{\sin 2t}{2} \right) dt = i\pi r^2. 
\end{aligned}
\end{equation*}

\end{enumerate}

{Second Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  Since the arc of the integral is the circle $|z|=r$, we see $z\bar{z}=r^2$ remains constant along the arc. 

\item  In the integral we can use the substitution, $$x=\frac{1}{2}(z+\bar{z})=\frac{1}{2}\left(z+\frac{r^2}{z}\right). $$

\item  Replace $x$ in the integral in terms of $z$, we have 
\begin{equation*}
\begin{aligned}
\int_{|z|=r} xdz &= \int_{|z|=r} \frac{1}{2}\left(z+\frac{r^2}{z}\right)dz 
= \frac{1}{2}\int_{|z|=r} zdz + \frac{r^2}{2}\int_{|z|=r} \frac{dz}{z} \\ 
&= \frac{r^2}{2}\cdot 2\pi i = r^2\pi i. 
\end{aligned}
\end{equation*}

\end{enumerate}

} % End of Solution. 
\fi


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%\newpage
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\item  %%4
(***) 
Compute $$\int_{|z|=1} |z-1|\cdot |dz|.$$ 


\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }

\begin{enumerate}[label={\arabic*.}]
\item  This is the line integral with respect to arc length. The definition is 
$$
\int_\gamma fds = \int_\gamma f|dz| = \int_a^b f(z(t))|z'(t)|dt. 
$$

\item  Let the unit circle $|z|=1$ be parametrised by $z=e^{it}, 0\le t\le 2\pi$. 

\item  Compute the integral, 
\begin{equation*}
\begin{aligned}
\int_{|z|=1} |z-1|\cdot |dz| &= \int_0^{2\pi} |\cos t + \sin t -1|\cdot |d e^{it}| \\ 
&= \int_0^{2\pi} |\cos t + i\sin t -1|\cdot |e^{it}i|dt \\ 
&= \int_0^{2\pi} |\cos t + i\sin t -1|dt \\
&= \int_0^{2\pi} \left\vert -2\cos^2\frac{t}{2} + i\cdot 2\sin \frac{t}{2} \cos \frac{t}{2} \right\vert dt \\ 
&= \int_0^{2\pi} \left\vert 2\cos\frac{t}{2} \right\vert dt \\ 
&= \int_{t=0}^{\pi} 4\cos\frac{t}{2} dt \\ 
&\overset{t=2u}{=} \int_{u=0}^{\pi/2} 8\cos u du =8.  
\end{aligned}
\end{equation*}

\end{enumerate}

} % End of Solution. 
\fi

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\end{enumerate}

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\newpage
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\section{Cauchy's Integral Formula}

\begin{enumerate}

\item  %%1

If the piecewise differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral 
$$\int_\gamma \frac{dz}{z-a}$$
is a multiple of $2\pi i$.

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{First Proof. }
 
\begin{enumerate}[label={\arabic*.}]
\item  The derivative of $\log (z-a)$ is $\frac{1}{z-a}$, thus we can compute the integral, 
\begin{equation*}
\begin{aligned}
\int_\gamma \frac{dz}{z-a} 
= \int_\gamma d \log (z-a) 
= \int_\gamma d \left[ \log|z-a| + i\mathrm{arg} (z-a) \right].
\end{aligned}
\end{equation*}

\item  Since $\gamma$ is a closed curve, $\log|z-a|$ returns to its initial value, thus 
$$
\int_\gamma d \log|z-a| =0. 
$$

\item  Also since $\gamma$ is a closed curve, $\mathrm{arg} (z-a)$ increases or decreases by a multiple of $2\pi$. 

\end{enumerate}

{Second Proof. }
 
\begin{enumerate}[label={\arabic*.}]
\item  Suppose the equation of $\gamma$ is $z=z(t), \alpha\le t\le \beta$. 

\item  Consider the function $$h(t) =\int_\alpha^t \frac{z'(t)}{z(t)-a}dt. $$

\item  The function $h(t)$ is defined and continuous on $\alpha\le t\le \beta$, and it has the derivative 
$$ h'(t) = \frac{z'(t)}{z(t)-a}. $$

\item  The derivative of the function $$\frac{z(t)-a}{e^{h(t)}}$$ is the constant zero, except at a finite number of points, since $\gamma$ is piecewise differentiable. 

\item  Since $h(\alpha)=0$, we have $$\frac{z(t)-a}{e^{h(t)}} = \frac{z(\alpha)-a}{e^{h(\alpha)}}  = z(\alpha)-a. $$

\item  Thus we have $$e^{h(t)} = \frac{z(\alpha)-a}{z(t)-a}. $$

\item  Since $\gamma$ is a closed curve, we have $z(\beta)=z(\alpha)$. Thus 
$$e^{h(\beta)} = \frac{z(\alpha)-a}{z(\beta)-a} =1. $$

\item  Thus $h(\beta)=2k\pi i$ for some integer $k$. 
{\color{blue} This integer $k$ is said to be the index of the point $a$ with respect to the curve $\gamma$. }

\end{enumerate}

} % End of Solution. 
\fi

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%\newpage
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\item  %%2

Suppose that $f(z)$ is analytic in an open disk $\Delta$, and let $\gamma$ be a closed curve in $\Delta$. For any point $a$ not on $\gamma$, 
$$
n(\gamma,a)\cdot f(a) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)dz}{z-a},
$$
where $n(\gamma,a)$ is the index of $a$ with respect to $\gamma$. 

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Proof. }
 
\begin{enumerate}[label={\arabic*.}]
\item  We apply Cauchy's theorem to the function $$F(z) = \frac{f(z)-f(a)}{z-a}. $$

\item  This function is analytic for $z\neq a$. 

\item  This function has a removable singularity at $z=a$, since $$\lim\limits_{z\to a} (z-a)F(z)=\lim\limits_{z\to a}  [f(z)-f(a)] = 0. $$

\item  The condition of Cauchy'w theorem is fulfilled; Thus for any closed curve $\gamma$ in the disk $\Delta$, we have 
$$\int_\gamma \frac{f(z)-f(a)}{z-a}dz =0. $$

\item  By the previous Lemma, there is an integer $n(\gamma,a)$, such that 
$$
 \frac{1}{2\pi i} \int_\gamma \frac{f(z)dz}{z-a} 
 =  \frac{1}{2\pi i} \int_\gamma \frac{f(a)dz}{z-a} 
 =  f(a)\cdot \frac{1}{2\pi i} \int_\gamma \frac{dz}{z-a} 
 = f(a)\cdot n(\gamma,a).
 $$ 
 
\end{enumerate}

} % End of Solution. 
\fi


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%\newpage
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\item  %%3
(***) 
Compute $$\int_{|z|=1} \frac{e^z}{z}\cdot dz.$$

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  Consider $f(z)=e^z$ and $a=0$. 

\item  By the previous Theorem 6, we have 
$$
\int_{|z|=1} \frac{e^z}{z}\cdot dz 
= \int_{|z|=1} \frac{f(z)}{z-0}\cdot dz 
= \int_{|z|=1} \frac{f(0)}{z-0}\cdot dz 
= f(0)\cdot 2\pi i \cdot n(\gamma, 0)
=e^0\cdot 2\pi i\cdot 1
= 2\pi i. 
$$

\end{enumerate}

} % End of Solution. 
\fi


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%\newpage
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\item  %%4
(***) 
Compute $$\int_{|z|=\rho} \frac{|dz|}{|z-a|^2}$$
under the condition $|a|\neq \rho$. 
Hint: make use of the equations $z\bar{z}=\rho^2$ and 
$$|dz|=-i\rho \frac{dz}{z}. $$

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  When $|z|=\rho$, we may write $z=\rho e^{it}$. 

\item  Thus $dz = \rho e^{it} idt =izdt$, and $|dz| = \rho dt = \rho \frac{dz}{iz} = -i\rho \frac{dz}{z}. $ 

\item  We have $|z-a|^2 = (z-a)(\bar{z}-\bar{a}) = z\bar{z}-a\bar{z}-\bar{a}z+a\bar{a}$. 

\item  Denote $r^2=a\bar{a}$; the integral is 
\begin{equation*}
\begin{aligned}
\int_{|z|=\rho} \frac{|dz|}{|z-a|^2}
&= \int_{|z|=\rho} \frac{-i\rho \frac{dz}{z}}{z\bar{z}-a\bar{z}-\bar{a}z+a\bar{a}} \\ 
&= \int_{|z|=\rho} \frac{-i\rho dz}{z(z\bar{z}-a\bar{z}-\bar{a}z+a\bar{a}) } \\ 
&= \int_{|z|=\rho} \frac{-i\rho dz}{\rho^2 z - a\rho^2 - \bar{a}z^2 + r^2z } \\
&= \int_{|z|=\rho} \frac{i\rho dz}{\bar{a}z^2 - (\rho^2+r^2) z + a\rho^2 }.
\end{aligned}
\end{equation*}

\item  The denominator is a quadratic polynomial; we compute its roots,
\begin{equation*}
\begin{aligned}
z= \frac{\rho^2+r^2 \pm \sqrt{(\rho^2+r^2)^2 - 4\bar{a}a\rho^2}} {2\bar{a}}
= \frac{\rho^2+r^2 \pm \sqrt{(\rho^2-r^2)^2}}{2\bar{a}}
= \frac{\rho^2+r^2 \pm |\rho^2-r^2| } {2\bar{a}}.
\end{aligned}
\end{equation*}

\item  When $r^2< \rho^2$, that is, when $a$ is inside the circle $|z|=\rho$, we have two roots 
\begin{equation*}
\begin{aligned}
z_1 &= \frac{2\rho^2}{2\bar{a}} = \frac{\rho^2}{\bar{a}} = \frac{\rho^2}{r^2} a, \\ 
z_2 &= \frac{2r^2}{2\bar{a}} = \frac{r^2}{\bar{a}} = a. 
\end{aligned}
\end{equation*}

\item  Since $|z_1|>\rho$, the function $f(z) := \frac{i\rho}{\bar{a}(z-z_1)}$ is analytic for $|z|<\rho+\varepsilon$ where $0<\varepsilon<|z_1|-\rho$. 

\item Since $|z_2|<\rho$, we apply Cauchy's integral formula and continue the computation, 
\begin{equation*}
\begin{aligned}
\int_{|z|=\rho} \frac{|dz|}{|z-a|^2}
&= \int_{|z|=\rho} \frac{i\rho dz}{\bar{a}(z-z_1)(z-z_2) } 
= 2\pi i \frac{i\rho}{\bar{a} (z_2-z_1)}
= \frac{2\pi \rho}{\rho^2-r^2}. 
\end{aligned}
\end{equation*}

\item  It is similar when $r^2>\rho^2$. 

\end{enumerate}

} % End of Solution. 
\fi


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%\newpage
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\item  %%5
(***) 
Suppose that $\varphi(\zeta)$ is continuous on the arc $\gamma$. 
Then the function 
$$
F_n(z) = \int_\gamma \frac{\varphi(\zeta)d\zeta}{(\zeta-z)^n}
$$
is analytic in each of the regions determined by $\gamma$, and its derivative is 
$F'_n(z) = nF_{n+1}(z)$.

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }

\begin{enumerate}[label={\arabic*.}]
\item  
\item  
\item  
\item  
\item  

\end{enumerate}

} % End of Solution. 
\fi


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%\newpage
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\item  %%6
(***) 
Compute $$\int_{|z|=1} e^zz^{-n}dz.$$%, \,\,\, \int_{|z|=2} z^n(1-z)^mdz, \,\,\,\int_{|z|=\rho} |z-a|^{-4}|dz|\,\, (|a|\neq \rho). $$

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  
\item  
\item  
\item  
\item  

\end{enumerate}

} % End of Solution. 
\fi

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Local Properties of Analytic Functions}

\begin{enumerate}

\item  %%1
(***) 
Suppose that $f(z)$ is analytic in the region $\Omega'$ obtained by omitting a point a from a region $\Omega$. A necessary and sufficient condition that there exist an analytic function in $\Omega$ which coincides with $f(z)$ in $\Omega'$ is that $$\lim\limits_{z\to a} (z - a)f(z) = 0.$$ The extended function is uniquely determined. 

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }

\begin{enumerate}[label={\arabic*.}]
\item  Necessity. 
\begin{enumerate}[label={\arabic*.}]
\item  Let $g(z)$ be an analytic function in $\Omega$ which coincides with $f(z)$ in $\Omega'=\Omega-\{a\}$. 
\item  Since the limit process does not involve the value of the function at the limiting point, we have 
$$\lim\limits_{z\to a} (z - a)f(z) = \lim\limits_{z\to a} (z - a)g(z) = 0\cdot g(a) =0. $$
\end{enumerate}

\item  Uniqueness. 
\begin{enumerate}[label={\arabic*.}]
\item  Let $g_1(z)$ and $g_2(z)$ be two analytic functions that extend $f(z)$ to $z=a$. 
\item  Since analytic functions are continuous, we get  
$$g_1(a) =  \lim\limits_{z\to a} g_1(z) = \lim\limits_{z\to a} g_2(z) = g_2(a). $$ 
\item  Thus the extended function is uniquely determined. 
\end{enumerate}

\item  Sufficiency. 
\begin{enumerate}[label={\arabic*.}]
\item  Consider some small circle $C=\{z : |z-a|=r \}\subset \Omega$. 
\item  Consider the complex integral with parameter $z$ inside $C$, 
$$g(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)d\zeta}{\zeta - z}. $$
\item  Use the condition $\lim\limits_{z\to a} (z - a)f(z) = 0$; by Cauchy's formula, we have $g(z)=f(z)$ for $0< |z-a|<r$. 
\item  The function $g(z)$ is analytic when $|z|<r$. 
\end{enumerate}

\end{enumerate}

} % End of Proof. 
\fi

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%\newpage 
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\item %%2 
(***) 
Let $f(z)$ be an analytic function in a region $\Omega$ containing $a$. 
Then 
\begin{enumerate}[label={(\alph*)}]
\item  There is an analytic function $f_n(z)$ such that 
$$
f(z) = f(a) + f'(a)(z-a)+\frac{1}{2!}f''(a)(z-a)^2 + \cdots + \frac{1}{(n-1)!}f^{(n-1)}(a)(z-a)^{n-1} + f_n(z)(z-a)^n. 
$$

\item  Let $C$ be a circle around $a$ so that $C$ and its inside are contained in $\Omega$. Then for $z$ inside the circle $C$, we have the following integral representation, 
$$
f_n(z) = \frac{1}{2\pi i} \int_C \frac{f(\zeta)d\zeta}{(\zeta-a)^n(\zeta-z)}. 
$$

\end{enumerate}

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  For $z\in \Omega-\{a\}$ we consider an analytic function 
$
F_1(z)=\frac{f(z)-f(a)}{z-a}.
$

\item  Since $f(z)$ is analytic thus continuous at $z=a$, we have 
$
\lim\limits_{z\to a} (z - a)F_1(z) = \lim\limits_{z\to a} [f(z)-f(a)] = 0; 
$
thus the analytic function $F_1(z)$ has a removable singularity at $z=a$. 

\item  By Theorem 7 the function $F_1(z)$ can be extended to {\color{blue}an analytic function $f_1(z)$ for $z\in \Omega$}. 

\item The value of $f_1(z)$ at $z=a$ is $f_1(a)=\lim\limits_{z\to a} F(z) = f'(a). $

\item  Thus for all $z\in \Omega$ we have 
\begin{equation}
f(z)=f(a) + (z-a)f_1(z). 
\label{eq-1}
\end{equation}

\item  Similarly for $z\in \Omega-\{a\}$ we consider an analytic function 
$F_2(z) = \frac{f_1(z)-f_1(a)}{z-a}. 
$

\item  Since $f_1(z)$ is analytic thus continuous at $z=a$, we have $$\lim\limits_{z\to a} (z - a)F_2(z) = \lim\limits_{z\to a} [f_1(z)-f_1(a)] = 0; $$ thus the analytic function $F_2(z)$ also has a removable singularity at $z=a$.  

\item  By Theorem 7 again, the function $F_2(z)$ can be extended to {\color{blue}an analytic function $f_2(z)$ for $z\in \Omega$}. 

\item  The value of $f_2(z)$ at $z=a$ is $f_2(a)=\lim\limits_{z\to a} F_2(z) = f_1'(a). $

\item  Thus for all $z\in \Omega$ we have 
\begin{equation}
f_1(z)=f_1(a) + (z-a)f_2(z). 
\label{eq-2}
\end{equation}

\item  Combine the Equation (\ref{eq-1}) and the Equation (\ref{eq-2}), for all $z\in \Omega$ we have 
\begin{eqnarray}
%\begin{aligned}
f(z) &=& f(a) + (z-a)f_1(z) \nonumber \\
&=& f(a) + (z-a)[f_1(a)+(z-a)f_2(z)] \nonumber \\
&=& f(a) + (z-a)f_1(a)+(z-a)^2f_2(z). \label{eq-3}
%\end{aligned}
\end{eqnarray}

\item  Take derivative twice in the Equation (\ref{eq-3}) and let $z=a$, we get $f''(a)=2f_2(a)$. 

\item  Similarly there exists an analytic function $f_3(z)$ such that for all $z\in \Omega$,  
\begin{eqnarray}
f(z) = f(a) + (z-a)f_1(a)+(z-a)^2f_2(a)+(z-a)^3f_3(z), 
\end{eqnarray}
and $f'''(a)=6f_3(a)$. 

\end{enumerate}

{Proof of (b).}
 
\begin{enumerate}[label={\arabic*.}]
\item  By Cauchy's formula, first we have 
$$
f_1(z) = \frac{1}{2\pi i} \int_C \frac{f_1(\zeta)d\zeta}{\zeta-z}. 
$$

\item  By the Equation (\ref{eq-1}), we have $f_1(z) = \frac{f(z)-f(a)}{z-a}$; Put this in the above integral, we get 
$$
f_1(z) = \frac{1}{2\pi i} \int_C \frac{\frac{f(\zeta)-f(a)}{\zeta-a}d\zeta}{\zeta-z} 
= \frac{1}{2\pi i} \int_C \frac{f(\zeta)d\zeta}{(\zeta-a)(\zeta-z)}.
$$

\item  To represent $f_2(z)$ by an integral, first by Cauchy's formula we have 
$$
f_2(z) = \frac{1}{2\pi i} \int_C \frac{f_2(\zeta)d\zeta}{\zeta-z}. 
$$

\item  By the Equation (\ref{eq-3}), we have $$f_2(z) = \frac{f(z)-f(a)-(z-a)f_1(a)}{(z-a)^2};$$ 
Put this in the above integral, we get 
\begin{equation*}
\begin{aligned}
f_2(z) &= \frac{1}{2\pi i} \int_C \frac{ \frac{f(\zeta)-f(a)-(\zeta-a)f_1(a)}{(\zeta-a)^2} d\zeta}{\zeta-z} \\ 
& = \frac{1}{2\pi i} \int_C \frac{ [ f(\zeta)-f(a)-(\zeta-a)f_1(a)] d\zeta}{(\zeta-a)^2(\zeta-z)} \\ 
& = \frac{1}{2\pi i} \int_C \frac{ f(\zeta) d\zeta}{(\zeta-a)^2(\zeta-z)}
- \frac{1}{2\pi i} \int_C \frac{ f(a) d\zeta}{(\zeta-a)^2(\zeta-z)}
- \frac{1}{2\pi i} \int_C \frac{ f_1(a) d\zeta}{(\zeta-a)(\zeta-z)}. 
\end{aligned}
\end{equation*}

\item  The last integral is zero because for all $a$ inside $C$, 
$$
\int_C \frac{d\zeta}{(\zeta-a)(\zeta-z)} = \frac{1}{a-z}\int_C \left( \frac{1}{\zeta-a} - \frac{1}{\zeta-z} \right) d\zeta
= \frac{1}{a-z}(2\pi i - 2\pi i) = 0. 
$$

\item  Take derivative with respect to $a$ in the previous identity, we see the middle integral is also zero, 
$$
\int_C \frac{d\zeta}{(\zeta-a)^2(\zeta-z)} = 0. 
$$


\end{enumerate}

} % End of Proof. 
\fi


\end{enumerate}

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\newpage 
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\section{The Calculus of Residues  }

\begin{enumerate}

\item %%1
Let $f(z)$ be analytic except for isolated singularities $a_j$ in a region $\Omega$.
Then
%(47)
$$
\frac{1}{2\pi i} \int_\gamma f(z)dz = \sum\limits_{j} n(\gamma,a_j)\mathrm{Res}_{z=a_j}f(z) 
$$
for any cycle $\gamma$ which is homologous to zero in $\Omega$ and does not pass through any of the points $a_j$.

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]

\item  Since $\gamma$ is a cycle in $\Omega$, there is only a finite number of isolated singular points $a_1,\cdots, a_n$ in $\Omega$ such that $n(\gamma,a_j)\neq 0$.  

\item  Draw a circle $C_j$ about $a_j$ of radius $\delta_j$, such that $|z-a_j|<\delta_j$ is contained in $\Omega$, and there is no singular point in the annulus $0<|z-a_j|<\delta_j$.  

\item  Denote $\Omega'=\Omega-\{a_1,\cdots,a_n\}$; Then inside $\Omega'$, the cycle $\gamma$ is simplified in the   homologous class,  
$$\gamma \sim \sum\limits_{j=1}^n n(\gamma,a_j)C_j. $$

\item  By Cauchy's theorem, the integral is simplified, 
$$
\frac{1}{2\pi i} \int_\gamma f(z)dz = \sum\limits_{j} \frac{1}{2\pi i} \int_{C_j} f(z)dz. 
$$

\item  Finally we need to prove 
$$
\frac{1}{2\pi i} \int_{C_j} f(z)dz = \mathrm{Res}_{z=a_j}f(z). 
$$

\item  By definition, {\color{blue}the residue of $f(z)$ at an isolated singularity $a_j$} is the unique complex number $R_j=\mathrm{Res}_{z=a_j}f(z)$ which makes 
$$f(z) - \frac{R_j}{z-a_j}$$ the derivative of a single-valued analytic function $F_j(z)$ in an annulus $0 < |z -a_j| < \delta_j$.

\item  Since the contour integral of a derivative function is always zero, we have  
$$
\frac{1}{2\pi i} \int_{C_j} \left[ f(z) - \frac{R_j}{z-a_j} \right] dz = \frac{1}{2\pi i} \int_{C_j} F'_j(z)dz =0. 
$$

\item  The proof reduces to the integral
$$
\frac{1}{2\pi i} \int_{C_j} \frac{1}{z-a_j} dz = 1. 
$$

\end{enumerate}

} % End of Solution. 
\fi

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%\newpage 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %%2
(***) 
Compute the definite integral 
$$
\int_0^\pi \frac{d\theta}{a+\cos\theta}, \,\,\,\, a>1. 
$$

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  
\item  
\item  
\item  
\item  

\end{enumerate}

} % End of Solution. 
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  %%3
(***) 
Compute the definite integral 
$$
\int_0^\pi \log\sin\theta d\theta. $$

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  
\item  
\item  


\end{enumerate}

} % End of Solution. 
\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %%4
If $f(z)$ is meromorphic in $\Omega$ with the zeros $a_j$ and the poles $b_k$, then
$$
\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)}dz = 
\sum\limits_{j} n(\gamma,a_j) - \sum\limits_{k} n(\gamma,b_k)
$$
for every cycle $\gamma$ which is homologous to zero in $\Omega$ and does not pass through any of the zeros or poles.

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]

\item  Since $\gamma$ does not pass through any of the poles of $f$, the image $\Gamma=f(\gamma)$ is a cycle in the complex plane. 

\item  Since $$\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)}dz = \frac{1}{2\pi i} \int_\Gamma \frac{dw}{w} =n(\Gamma, 0),$$ 
the argument principle says that 
$$
n(\Gamma, 0) = \sum\limits_{j} n(\gamma,a_j) - \sum\limits_{k} n(\gamma,b_k). 
$$

\item  An analytic function $f(z)$ in a region $\Omega$ is said to be {\color{blue}meromorphic} in $\Omega$, if for each $a\in \Omega$, there is a neighbourhood $|z-a|<\delta$, contained in $\Omega$, such that either $f(z)$ is analytic in the whole neighbourhood, or else $f(z)$ is analytic for $0<|z-a|<\delta$, and the isolated singularity is a pole. 

\item  

\item  


\end{enumerate}

} % End of Solution. 
\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  %%5
(***) 
Let $\gamma$ be homologous to zero in $\Omega$ and such that $n(\gamma,z)$ is either 0 or 1 for any point $z$ not on $\gamma$. Suppose that $f(z)$ and $g(z)$ are analytic in $\Omega$ and satisfy the inequality $|f(z)-g(z)| < |f(z)|$ on $\gamma$. Then $f(z)$ and $g(z)$ have the same number of zeros enclosed by $\gamma$. 

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  
\item  
\item  
\item  
\item  

\end{enumerate}

} % End of Solution. 
\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %%6
(***) 
How many roots does the equation 
$z^7 - 2z^5 + 6z^3 - z + 1 = 0$ 
have in the disk $|z| < 1$? 
Hint: Look for the biggest term when $|z|=1$ and apply Rouche's theorem. 

\ifnum\showsolution=1

{\color{red} %Begin of Solution. 
{Solution. }
 
\begin{enumerate}[label={\arabic*.}]
\item  
\item  
\item  
\item  
\item  

\end{enumerate}

} % End of Solution. 
\fi

\end{enumerate}


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\end{document}
